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ivanzaharov [21]
3 years ago
11

A toy manufacturer uses 51,450 rubber wheels per year for its popular dump truck series. The firm makes its own wheels, which it

can produce at a rate of 800 per day. The toy trucks are assembled uniformly over the entire year. Carrying cost is $1.90 per wheel per year. Setup cost for a production run is $43. The firm operates 245 days per year. Determine the following:
Business
1 answer:
masha68 [24]3 years ago
6 0

Answer:

Requirement <em>"a. Optimal run size (Round your answer to a whole number, following normal rules of rounding.) EPQ b. Use your final answer from part a to determine minimum total annual cost for carrying and setup. (Round your answer to a whole number.) Total Annual Inventory Cost"</em>

<em />

a. Annual Demand D = 51450

Daily demand d = 210 {51450 / 245}

Daily production p = 800

Carrying cost per wheel per year H = $1.90

Setup cost S = $43

Optimal run size (EPQ) = √((2*D*S) / (H*(1 - (d/p))))

Optimal run size (EPQ) = √((2*51450*43) / (1.90*(1 - (210/800))))

Optimal run size (EPQ) = √((2*51450*43) / (1.90*0.7375))

Optimal run size (EPQ) = √(4424700 / 1.40125)

Optimal run size (EPQ) = √3157680.64

Optimal run size (EPQ) = 1776.986

Optimal run size (EPQ) = 1,777

b. Total Annual Setup cost = (D*S) / EPQ

Total Annual Setup cost = (51450*43) / 1777

Total Annual Setup cost = $1,244.99

Total Annual Carrying cost = ((H*EOQ)/2) * (1-(d/p))

Total Annual Carrying cost = ((1.90*1777)/2) * (1-(210 / 800))

Total Annual Carrying cost = ((1.90*1777)/2) * 0.7375

Total Annual Carrying cost = $1,688.15 * 0.7375

Total Annual Carrying cost = $1245.010625

Total Annual Carrying cost = $1,245.01

Total Minimum Annual cost for Carrying and Setup = Total Annual Carrying cost + Total Annual Setup cost

= $1245.01 + $1244.99

= $2,490

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8 0
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with the new rate we will pay in 58 months.

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Currently we owe 10,000

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We are going to do monthly payment of 200 dollars each month

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C \times \frac{1-(1+r)^{-time} }{rate} = PV\\

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200 \times \frac{1-(1+0.0051667)^{-n} }{0.0051667} = 10000\\

We arrenge the formula and solve as muhc as we can:

(1+0.0051667)^{-n}= 1-\frac{10000\times0.0051667}{200}

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